\subsection{Breaking an interval}
Let \(f\) be integrable on \([a, b]\).
If \(a < c < b\), then \(f\) is integrable over \([a, c]\) and \([c, b]\), with
\[
	\int_a^b f = \int_a^c f + \int_c^b f
\]
Conversely, if \(f\) is integrable on \([a, c]\) and \([c, b]\), then \(f\) is integrable over \([a, b]\) and the same equality holds for the combination of the integrals.
\begin{proof}
	We first make two observations.
	First, if \(\mathcal D_1\) is a dissection of \([a, c]\) and \(\mathcal D_2\) is a dissection of \([c, b]\), then \(\mathcal D = \mathcal D_1 \cup \mathcal D_2\) is a dissection of \([a, b]\), and
	\begin{equation}
		S(f, \mathcal D_1 \cup \mathcal D_2) = S\qty(\eval{f}_{[a, c]}, \mathcal D_1) + S\qty(\eval{f}_{[c, b]}, \mathcal D_2)
		\tag{\(\ast\)}
	\end{equation}
	Also, if \(\mathcal D\) is a dissection of \([a, b]\), then
	\begin{equation}
		S(f, \mathcal D) \geq S(f, \mathcal D \cup \{ c \}) = S\qty(\eval{f}_{[a, c]}, \mathcal D_1) + S\qty(\eval{f}_{[c, b]}, \mathcal D_2)
		\tag{\(\dagger\)}
	\end{equation}
	Now,
	\[
		(\ast) \implies I^\star(f) \leq I^\star\qty(\eval{f}_{[a, c]}) + I^\star\qty(\eval{f}_{[c, b]})
	\]
	Further,
	\[
		(\dagger) \implies I^\star(f) \geq I^\star\qty(\eval{f}_{[a, c]}) + I^\star\qty(\eval{f}_{[c, b]})
	\]
	Hence,
	\[
		I^\star(f) = I^\star\qty(\eval{f}_{[a, c]}) + I^\star\qty(\eval{f}_{[c, b]})
	\]
	This argument also applies for the lower integral, therefore
	\[
		0 \leq I^\star(f) - I_\star(f) = \underbrace{I^\star\qty(\eval{f}_{[a, c]}) - I_\star\qty(\eval{f}_{[a, c]})}_{A} + \underbrace{I^\star\qty(\eval{f}_{[c, b]}) + I_\star\qty(\eval{f}_{[c, b]})}_{B}
	\]
	Note that \(A, B \geq 0\).
	If \(f\) is integrable on \([a, c]\) and \([c, b]\), then \(A = B = 0\), hence \(I^\star(f) = I_\star(f)\) and it is integrable on \([a, b]\).
	If \(f\) is integrable on \([a, b]\), then we know \(I^\star(f) = I_\star(f)\), so \(A = B = 0\) so \(f\) is integrable on \([a, c]\) and \([c, b]\).
\end{proof}
Note that we take the following convention:
\[
	\int_a^b f = -\int_b^a f
\]
and if \(a=b\), then this value is zero.
With this convention, if \(f\) is bounded with \(\abs{f} \leq k\), then
\[
	\abs{\int_a^b f} \leq k\abs{b - a}
\]

\subsection{Fundamental theorem of calculus}
Suppose a function \(f \colon [a, b] \to \mathbb R\) is bounded and integrable.
Then since it is integrable on any sub-interval, we can define
\[
	F(x) = \int_a^x f(t) \dd{t}
\]
for \(x \in [a, b]\).
\begin{theorem}
	\(F\) is continuous.
\end{theorem}
\begin{proof}
	We know that
	\[
		F(x + h) - F(x) = \int_x^{x+h} f(t)\dd{t}
	\]
	We want this quantity to vanish as \(h \to 0\).
	We find, given that \(f\) is bounded by \(k\),
	\[
		\abs{F(x+h) - F(x)} = \abs{\int_x^{x+h} f(t) \dd{t}} \leq k\abs{h}
	\]
	So the result follows as \(h \to 0\).
\end{proof}
\begin{theorem}
	If in addition \(f\) is continuous at \(x\), then \(F\) is differentiable, with \(F'(x) = f(x)\).
\end{theorem}
\begin{proof}
	Consider
	\[
		\abs{\frac{F(x + h) - F(x)}{h} - f(x)}
	\]
	If this tends to zero, then the theorem holds.
	\[
		\abs{\frac{F(x + h) - F(x)}{h} - f(x)} = \frac{1}{\abs{h}} \abs{\int_x^{x+h} f(t) \dd{t} - hf(x)} = \frac{1}{\abs{h}} \abs{\int_x^{x+h} [f(t) - f(x)] \dd{t}}
	\]
	Since \(f\) is continuous at \(x\), given \(\varepsilon > 0\), \(\exists \delta > 0\) such that \(\abs{t - x} - \delta \implies \abs{f(t) - f(x)} < \varepsilon\).
	If \(\abs{h} < \delta\), then the integrand is bounded by \(\varepsilon\).
	Hence,
	\[
		\abs{\frac{F(x + h) - F(x)}{h} - f(x)} \leq \frac{1}{\abs{h}} \abs{h\varepsilon} = \varepsilon
	\]
	So we can make this value as small as we like.
	So the theorem holds.
\end{proof}
For example, consider the function
\[
	f(x) = \begin{cases}
		-1 & x \in [-1, 0] \\
		1  & x \in (0, 1]
	\end{cases}
\]
This is a bounded, integrable function, with
\[
	F(x) = -1 + \abs{x}
\]
Note that this \(F\) is not differentiable at \(x = 0\).
\begin{corollary}
	If \(f = g'\) is a continuous function on \([a, b]\), then
	\[
		\int_a^x f(t) \dd{t} = g(x) - g(a)
	\]
	is a differentiable function on \([a, b]\).
\end{corollary}
\begin{proof}
	From above, \(F - g\) has zero derivative in \([a, b]\), hence \(F - g\) is constant.
	Since \(F(a) = 0\), we get \(F(x) = g(x) - g(a)\).
\end{proof}
Note that every continuous function \(f\) has an `indefinite' integral (or `antiderivative') written \(\int f(x)\dd{x}\), which is determined uniquely up to the addition of a constant.
Note further that we have now essentially solved the differential equation
\[
	\left\{\begin{array}{l}
		y'(x) = f(x) \\
		y(a) = y_0
	\end{array}\right.
\]
and shown that there is a unique solution to this ordinary differential equation.
